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Leokris [45]
3 years ago
11

What is the specific heat of the solid phase? (Please see picture attached)

Chemistry
1 answer:
ValentinkaMS [17]3 years ago
4 0

Answer:

B.0.2 J/g°C

Explanation:

From the attached picture;

  • Heat attained in the solid phase is 200 Joules
  • Change in temperature is 50°C ( from 0°C to 50°C)
  • Mass of the solid is 20 g

We are required to determine the specific heat capacity of the substance;

  • We need to know that Quantity of heat is given by the product of mass,specific heat capacity and change in temperature.
  • That is; Q = mcΔT

Rearranging the formula;

c = Q ÷ mΔT

Therefore;

Specific heat = 200 J ÷ (20 g × 50°c)

                      = 0.2 J/g°C

Thus, the specific heat of the solid is 0.2 J/g°C

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Which aqueous solution has the lowest freezing point c6h12o6, c2h5oh, ch3cooh, or nacl?
Eddi Din [679]
Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates  only partially. 

So, most particles of the solute is going to be in the solution of NaCl, 
so<span> the lowest freezing point has the aqueous solution of NaCl.</span>
3 0
3 years ago
What is the net amount of energy released when one mole of h2o(?) is produced?
german
The net amount of energy produced can be obtained from a table of enthalpy change of formation, available online.

The enthalpy change of formation indicate how much energy the 1 mole of the product (H2O) has relative to the elemental reactants (H2 and O2).  In other words, the "lost" energy equals the heat/energy released. 

For water (H2O), this value is -285.8 if the final product is a liquid under standard conditions, and -241.82 if the product is in gas form which contains some energy that could be further released.  This means that if the final product (H2O) is in liquid form, energy released is 285.8 kJ/mol.

Since water is in liquid form under standard conditions, the first value (285.8 kJ/mol) is generally appropriate.
7 0
3 years ago
. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r
Tom [10]

Answer :  The energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

Explanation : Given,

Wavelength = 435.8nm=435.8\times 10^{-9}m

conversion used : 1nm=10^{-9}m

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength = 435.8\times 10^{-9}m

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}

E=4.56\times 10^{-19}J

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

3 0
3 years ago
Question 2
QveST [7]
Metals only and ionic
3 0
3 years ago
Read 2 more answers
6. An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of acid rain on limestone-rich soils. To p
pishuonlain [190]

Answer:

[Na₂CO₃] = 0.094M

Explanation:

Based on the reaction:

HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)

It is possible to find pH using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA]

Where [A⁻] is concentration of conjugate base,  [CO₃²⁻] = [Na₂CO₃] and  [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.

pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>

<em />

Replacing these values:

10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]

<em> [Na₂CO₃] = 0.094M</em>

<em />

5 0
3 years ago
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